第五届华中区程序设计邀请赛暨武汉大学第十四届校赛 网络预选赛 A Minimum Sum 树状数组

Problem 1603 - Minimum Sum

 

Description

There are n numbers A[1] , A[2] .... A[n], you can select m numbers of it A[B[1]] , A[B[2]] ... A[B[m]]  ( 1 <= B[1] < B[2] .... B[m] <= n ) such that Sum as small as possible.

Sum is sum of abs( A[B[i]]-A[B[j]] ) when 1 <= i < j <= m.

Input

There are multiple test cases.
First line of each case contains two integers n and m.( 1 <= m <= n <= 100000 )
Next line contains n integers A[1] , A[2] .... A[n].( 0 <= A[i] <= 100000 )
It's guaranteed that the sum of n is not larger than 1000000.

Output

For each test case, output minimum Sum in a line.

Sample Input

4 2
5 1 7 10
5 3
1 8 6 3 10

Sample Output

2
8

 

题意：

 

　　给你n个数，让你选M个数来，使得 sigma abs( A[B[i]]-A[B[j]] ) when 1 <= i < j <= m.最小

 

题解：

    

    排序过后，就是选择连续的m个数了

　 由于数的范围小，我们在统计连续和的时候 利用BIT优化

 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int N = 1e5+10, M = 50, mod = 1e9+9, inf = 1e9+9;
typedef long long ll;

ll sum;

int n,m,a[N];
  int C[N],A[N];
    int lowbit(int x) {
        return x&(-x);
    }
    void add(int x, int add) {
        for (; x < N; x += lowbit(x)) {
            C[x] += add;
        }
    }
    int sums(int x) {
        int s = 0;
        for (; x > 0; x -= lowbit(x)) {
          s+=C[x];
        }
        return s;
    }
int main() {
    while(scanf("%d%d",&n,&m)!=EOF) {
            memset(C,0,sizeof(C));
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        sort(a+1,a+n+1);
        sum = 0;
        for(int i=1;i<=m;i++) {
            sum+=(a[i]*(i-1)-sums(i));
            add(i,a[i]);
        }
  // cout<<sum<<endl;
        ll ans = sum;
        for(int i=m+1;i<=n;i++) {
            sum -= (sums(i)-sums(i-m)-a[i-m]*(m-1));
          // cout<<(sums(i)-sums(i-m+1)-a[i-m]*(m-1))<<endl;
          //  cout<<sum<<endl;
            sum += (a[i]*(m-1)-sums(i)+sums(i-m));
            add(i,a[i]);
            ans = min(ans,sum);
        }
        printf("%lld\n",ans);
    }
    return 0;
}